Consider the reaction a + b c +

1 answer:

6 0

When the reaction is at equilibrium ∴ΔG = Zero
now we need to get Δ G°, we can get it from this formula:
ΔG° = - RT㏑K(eq)
when we have R = 8.315 J/mol°K and T in Kelvin = 25+273 = 298
so we need to get the K(eq) from this formula:
K(eq) = concentration of products / concentration of reactants
= [ C] [ D] / [A] [B]
= (0.8*0.8) / (0.2*0.2) = 16
by substitution in ΔG° formula:
∴ ΔG° = - 8.315 * 298 *㏑(16) = -6870 Jmol^-1 /1000 = - 6.87 KJ mol^-1

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Answer:

reciprocal

Explanation:

The equilibrium constant for an equilibrium reaction is the ratio of the equilibrium constant for the forward reaction, Kf, to the equilibrium constant for the reverse reaction, Kr:

For A ⇄ B

The forward reaction is: A → B, with rate constant is Kf = K₁ and
The reverse reaction is: B → A, with rate constant Kr = K₂

$K_{eq}=\dfrac{K_f}{K_r}=\dfrac{K_1}{K_2}$

When you write the reaction in the other reaction, the forward and the reverse reaction are exchanged:

For B ⇄ A

The forward reaction is B → A, with rate constant Kf = K₂
The reverse reaction is A → B, with rate constant Kr = K₁

$K'_{eq}=\dfrac{K_f}{K_r}=\dfrac{K_2}{K_1}$

As you see:

$K'_{eq}=\dfrac{1}{K_{eq}}$

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